In the Reaction: Pb(NO3)2 + 2 Kl —> Pbl2 + 2 KNO3 How many grams of lead (II) iodide, Pbl2, is produced from 82.7 grams of potassium iodide, KI?

Round to two decimal places.

114.8g PbI2 are produced

Explanation:

To solve this question we must convert the mass of KI to moles using its molar mass (KI = 166.0g/mol). As 2 moles of KI produce 1 mole of PbI2 we can find its moles and its mass (Molar mass PbI2: 461.01g/mol) as follows:

Moles KI:

82.7g KI * (1mol / 166.0g) = 0.498 moles KI

Moles PbI2:

0.498 moles KI * (1mol PbI2 / 2mol KI) = 0.249 moles PbI2

Mass PbI2:

0.249moles * (461.01g/mol) =

114.8g PbI2 are produced

The air is going to come out because of the low pressure.

Answer is: the equilibrium  shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -